University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.2
Renormalised QED: Z₁, Z₂, Z₃, δₘ and on-shell conditions
Unit 6 taught the habit of writing a theory in the parameters you measure. QED is where it pays: four counterterms, four conditions an experiment can check, and one structural surprise, that the electron's own dressing cancels out of its charge. Every loop in Units 12 to 14 is booked against this ledger.
Build the model
Connect the measurement to the mechanism.
Renormalised QED is the bare QED Lagrangian rewritten in the electron's pole mass and the charge of the Thomson limit, with everything the regulator controls pushed into counterterm vertices; topic 6.06 supplied that logic, and nothing here changes it. What is new is the structure. Rescaling ψ₀ = √Z₂ ψ and A₀ = √Z₃ A, and defining Z₁ by e₀Z₂√Z₃ = Z₁e and δₘ by Z₂m₀ − m, produces exactly four constants, because power counting leaves three divergent functions and the electron's carries two Dirac structures, p̸ and 1.
Gauge invariance shapes the rest: the photon counterterm comes out transverse, there is no photon mass to fix, and the photon's single condition sets a residue, not a pole. The electron takes two conditions, a pole at 0.511 MeV with residue 1, and the vertex one, Γμ = γμ at zero momentum transfer. The Ward–Takahashi identity then gives Z₁ = Z₂, and e = √Z₃ e₀ follows: the photon alone renormalises charge, equally for every particle.
The scheme has edges. It needs a gauge-invariant regulator, because a momentum cutoff manufactures a photon mass; Z₁ and Z₂ are infrared-divergent and gauge-dependent and are never observables; and the on-shell α is the natural expansion parameter only near q² = 0, since at large momentum transfer the logarithms in Π grow and Unit 13 trades it for a running coupling.
- Simple definition
- A QED renormalisation condition defines a measured quantity and so fixes one counterterm: the electron propagator has its pole at the measured mass with residue 1, the photon propagator has residue 1 at q² = 0, and the vertex equals γμ at zero momentum transfer.
- Example
- Cut the electron loop off with a Pauli–Villars mass Λ = 100 GeV and Π(0) = 0 gives δ₃ = −0.0189, so Z₃ = 0.981 and the bare 1/α₀ = 134.5, while the measured 1/α stays 137.036, the value the Thomson limit returns.
Derived and exact: added to the QED Lagrangian written in m and e it reproduces the bare one term for term. Every δ vanishes in the free limit e → 0.
δᵢ = Zᵢ − 1 from ψ₀ = √Z₂ψ, A₀ = √Z₃A, e₀Z₂√Z₃ = Z₁e; δₘ = Z₂m₀ − m in MeV; δ₁, δ₂, δ₃ and e pure numbers
Derived. Contract the photon vertex with qμ and it vanishes: transverse, it can rescale the photon propagator but can never give the photon a mass.
p the electron four-momentum and q the photon's, both in MeV; δₘ in MeV, the other δ pure numbers
Stated, as definitions of m and of ψ's normalisation. At one loop δ₂ = Σ₂′(m) and δₘ = mδ₂ − Σ₂(m), derived, so the self-mass m − m₀ = Σ₂(m) is not δₘ.
−iΣ sums the 1PI electron self-energy, counterterm included; m = 0.51099895 MeV, the measured pole mass
Stated. Transversality keeps the pole at q² = 0 unaided (topic 12.09); this one condition fixes only the residue, and with it the size of e.
Πμν = (q²gμν − qμqν)Π(q²), counterterm included; q in MeV; Π, Π₂ and δ₃ pure numbers
Stated. It makes e the charge an on-shell electron shows a static, long-wavelength field, the Thomson limit's. F₂(0) is left free: g − 2 is predicted in Unit 14.
Γμ = γμF₁(q²) + (iσμν qν/2m)F₂(q²) with q = p′ − p in MeV; F₁ and F₂ pure numbers
Derived from the quoted Z₁ = Z₂. With the Pauli–Villars form of Unit 13's Z₃ = 1 − (α/3π)ln(Λ²/m²) and Λ at the Planck mass, Z₃ = 0.920 and 1/α₀ = 126.1.
from e₀Z₂√Z₃ = Z₁e and Z₁ = Z₂ (topic 12.07, quoted); e, e₀ and α = e²/4π pure numbers in four dimensions
From the bare Lagrangian to four counterterms
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 6.06's logic carries over unchanged: rescale the bare fields, write the Lagrangian in measured parameters, and let the difference become vertices fixed by conditions. The bare QED Lagrangian is ℒ₀ = −¼(F₀)² + ψ̄₀(i∂̸ − m₀)ψ₀ − e₀ψ̄₀A̸₀ψ₀ plus gauge fixing, with (F)² = FμνFμν. Set ψ₀ = √Z₂ψ and A₀ = √Z₃A. The kinetic terms acquire Z₂ and Z₃, the mass term Z₂m₀ and the coupling e₀Z₂√Z₃. Define the measured charge e by e₀Z₂√Z₃ = Z₁e, the mass counterterm by δₘ = Z₂m₀ − m, and write Zᵢ = 1 + δᵢ. Then ℒ₀ is the QED Lagrangian in m and e plus ℒct = −¼δ₃(F)² + ψ̄(iδ₂∂̸ − δₘ)ψ − eδ₁ψ̄A̸ψ, derived and exact. Dimensions: [ψ] = 3/2 and [A] = 1, so δₘ is a mass while δ₁, δ₂, δ₃ and e are pure numbers. Free limit: at e = 0 nothing dresses anything, so δ₁, δ₂, δ₃ and δₘ all vanish and every Z returns to 1.
Why four constants, and why these four
Topic 12.01's power counting, D = 4 − B − (3/2)F, quoted, leaves three one-particle-irreducible functions whose divergences survive the symmetries: the electron self-energy, the photon self-energy and the vertex. Furry's theorem removes the one- and three-photon functions and gauge invariance makes light-by-light finite. Each surviving function needs one constant per structure its divergence can take. The electron self-energy has D = 1, so its divergent part is a + bp̸: two structures, matched by δₘ and δ₂. The photon's has D = 2, but gauge invariance allows only (q²gμν − qμqν) times a logarithm: one structure, δ₃. The vertex has D = 0, a constant times γμ: one structure, δ₁. Topic 3.07's rule, i times the coefficient in ℒct, gives i(p̸δ₂ − δₘ) on an electron line, −i(gμνq² − qμqν)δ₃ on a photon line and −ieγμδ₁ at a vertex, derived. Contract the photon one with qμ and it vanishes: the counterterm is transverse, so it can rescale the photon propagator but cannot move its pole.
Two conditions on the electron
Write the full propagator as i/(p̸ − m − Σ(p̸)), with −iΣ the one-particle-irreducible electron diagrams, counterterm included. Near the pole Σ(p̸) ≈ Σ(m) + (p̸ − m)Σ′(m), with ′ meaning d/dp̸, so the propagator is i/[(p̸ − m)(1 − Σ′(m)) − Σ(m)]. Its pole sits at the measured 0.51099895 MeV if Σ(m) = 0 and its residue is 1 if Σ′(m) = 0: stated, as the definitions of m and of ψ's normalisation. The bare field then has residue Z₂, and LSZ for ψ needs no √Z₂. At one loop Σ = Σ₂ − p̸δ₂ + δₘ, so δ₂ = Σ₂′(m) and δₘ = mδ₂ − Σ₂(m), derived. Mind the bookkeeping: the self-mass m − m₀ is Σ₂(m), while δₘ also carries mδ₂. Quoted, in Feynman gauge with a Pauli–Villars regulator mass Λ and a photon mass λ, in the standard textbook forms (topics 12.03 to 12.05 derive the same logarithms in dimensional regularisation, where Σ₂(m) carries 4/3 in place of ½, because a finite part belongs to its regulator): Σ₂(m) = (3α/4π)m[ln(Λ²/m²) + ½], free of λ, but δ₂ = −(α/4π)[ln(Λ²/m²) + 2ln(λ²/m²) + 9/2], infrared-divergent, and δₘ inherits the divergence.
One condition on the photon, and it is not a mass
With its counterterm the photon self-energy is iΠμν = i(q²gμν − qμqν)Π(q²); that it is transverse is proved in topic 12.09 and quoted here. Summing the chain in Feynman gauge gives −igμν/(q²[1 − Π(q²)]) plus qμqν terms that vanish against conserved currents. The pole stays at q² = 0 provided Π is regular there, which holds because every charged particle is massive. No condition is needed for it, and no counterterm could supply one, since ℒ₀ has no A₀² term to rescale. One condition remains, Π(0) = 0, the statement that the residue is 1, so at one loop δ₃ = Π₂(0). Quoted in its Pauli–Villars form for the electron loop, with a regulator mass Λ (Unit 13 computes the same logarithm in dimensional regularisation): Π₂(0) = −(α/3π)ln(Λ²/m²), so δ₃ is negative and Z₃ lies below 1. Gauge fixing needs nothing new: with ξ₀ = Z₃ξ the term (∂⋅A₀)²/2ξ₀ equals (∂⋅A)²/2ξ, and since topic 12.09 leaves the longitudinal propagator uncorrected, no counterterm is required there, stated.
The vertex condition defines the charge you measure
Between on-shell spinors the vertex has two form factors, Γμ = γμF₁(q²) + (iσμνqν/2m)F₂(q²), with q = p′ − p. The condition Γμ = γμ at q = 0 says F₁(0) = 1, stated, and at one loop it sets δ₁ equal to minus the loop's own contribution to F₁ at q = 0. The σμνqν term vanishes at q = 0, so the condition is silent about F₂(0): the anomalous moment α/2π is a prediction, checked in Unit 14. Why this point? Scatter an electron off a heavy static source at small momentum transfer. The amplitude carries e²F₁(q²)/(q²[1 − Π(q²)]), which with F₁(0) = 1 and Π(0) = 0 tends to its tree value e²/q²: e is the charge a long-wavelength probe sees, and so the one in topic 10.07's Thomson limit, σ = 8πα²/(3m²). With α = 1/137.036 and the pole mass, restoring ħc gives (8π/3)(αħc/mc²)² = 0.6652 b, the tabulated σₑ: not an independent measurement, but the limit that topic 10.07 makes exact once α and m are the measured ones. Dimensions check: αħc/mc² is a length, 2.818 fm.
Why the charge needs only Z₃
Solve the definition for the measured charge: e = e₀Z₂√Z₃/Z₁. Three dressings could enter, the electron field's, the photon field's and the vertex's. The Ward–Takahashi identity, taken to zero momentum transfer in topic 12.07 and quoted here, gives Z₁ = Z₂ to all orders, so e = √Z₃e₀ and α = Z₃α₀, derived. The factors that cancel are exactly the gauge-dependent, infrared-divergent ones, so the charge carries neither, and what survives belongs to the photon, which every charged particle shares: a muon or a proton, however differently dressed, has its charge renormalised by the same √Z₃. That is why the measured |qₚ + qₑ|/e, below 10⁻²¹, needs no fine-tuning. Numbers, electron loop only: with Λ at the Planck mass, 1.22 × 10¹⁹ GeV, ln(Λ²/m²) = 103.1, Z₃ = 0.920 and 1/α₀ = 126.1, a bare charge 4.2 per cent above e. Z₃ reaches zero only at ln(Λ/m) = 3π/2α, near Λ/m = 10²⁸⁰, the Landau pole, where one-loop Z₃ stops meaning anything.
Change one variable at a time
Make the relationship visible.
Slide the photon mass across ten decades: the δ₁ and δ₂ bars move in lockstep and drag δₘ with them, while δ₃ and e₀/e − 1 stand still, so the charge never sees the infrared. Then push Λ to log₁₀(Λ/m) = 22, near the Planck mass: Z₃ falls to 0.92.
Z₁ = Z₂1.0188
Z₃0.9643
BARE 1/α₀132.15
SELF-MASS (m − m₀)/m0.0811
Live interpretationZ₁ = Z₂: 1.0188. Z₃: 0.9643. BARE 1/α₀: 132.15. SELF-MASS (m − m₀)/m: 0.0811
Catch the common trap
Explain before calculating.
A one-loop on-shell renormalisation of QED, in some covariant gauge and with a gauge-invariant regulator, returns δ₁ = δ₂ = 0.0050 and δ₃ = −0.0360. To first order in α, how does the bare charge e₀ compare with the measured charge e?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyKeep only the electron loop and cut it off with a Pauli–Villars mass Λ = 1.0 TeV, so that Π₂(0) = −(α/3π)ln(Λ²/m²), the form Unit 13's result takes with this regulator, quoted. Use the photon's on-shell condition to find δ₃ and Z₃ and, with Z₁ = Z₂ from topic 12.07, the bare 1/α₀. Then evaluate the Thomson cross-section 8πα²/(3m²) with the measured α = 1/137.036 and with α₀, and compare both with the tabulated σₑ = 0.6652 b, which topic 10.07 makes exact with the physical α and m.
- With its counterterm the photon self-energy is Π(q²) = Π₂(q²) − δ₃, so Π(0) = 0 gives δ₃ = Π₂(0): the counterterm takes the whole zero-momentum value of the loop.
- Λ/m = 1.0 × 10⁶ MeV ÷ 0.51099895 MeV = 1.957 × 10⁶, so ln(Λ²/m²) = 2 × 14.487 = 28.974. With α/3π = 7.743 × 10⁻⁴, δ₃ = −0.02243 and Z₃ = 0.97757.
- With Z₁ = Z₂, e = √Z₃e₀, so α₀ = α/Z₃ and 1/α₀ = 137.036 × 0.97757 = 133.96: the bare charge is the larger, screened by virtual pairs.
- Restore ħc: αħc/mc² = 197.327 MeV fm ÷ (137.036 × 0.51099895 MeV) = 2.8179 fm, and σ = (8π/3)(2.8179 fm)² = 66.52 fm² = 0.6652 b, the tabulated σₑ. With α₀ instead, σ grows by (137.036/133.96)² = 1.046, to 0.6961 b.
- Checks: δ₃ is a pure number, as the coefficient of (F)² must be, and vanishes as α → 0; σ is an area. Lowering Λ to 10 GeV moves δ₃ to −0.0153 and 1/α₀ to 134.9, while α and σ do not move at all: only the bare charge knows the regulator.
Answerδ₃ = −0.0224, Z₃ = 0.978 and 1/α₀ = 134.0; the Thomson cross-section is 0.6652 b with α, the tabulated σₑ, but would be 0.696 b with α₀.
MediumIn Feynman gauge with a Pauli–Villars mass Λ = 1.0 TeV and a photon mass λ = 1.0 eV, the standard results with this regulator are Σ₂(m) = (3α/4π)m[ln(Λ²/m²) + ½] and δ₂ = −(α/4π)[ln(Λ²/m²) + 2ln(λ²/m²) + 9/2], quoted (topics 12.03 to 12.05 derive the same logarithms in dimensional regularisation, with different constants). Impose the two electron conditions to find δ₂, δₘ and the bare mass m₀ to first order in α. Repeat for λ = 1.0 meV and say which numbers moved, and why.
- The conditions give δ₂ = Σ₂′(m) and δₘ = mδ₂ − Σ₂(m). Since δₘ = Z₂m₀ − m, to first order m₀ = m + δₘ − mδ₂ = m − Σ₂(m).
- ln(Λ²/m²) = 28.974, so Σ₂(m)/m = 1.7421 × 10⁻³ × 29.474 = 0.05135 and Σ₂(m) = 0.02624 MeV. Hence m₀ = 0.51100 − 0.02624 = 0.48476 MeV.
- λ = 1.0 eV: ln(λ²/m²) = −26.288, the bracket is 28.974 − 52.577 + 4.5 = −19.103, and δ₂ = −5.807 × 10⁻⁴ × (−19.103) = +0.01109. Then δₘ = m(0.01109 − 0.05135) = −0.02057 MeV.
- λ = 1.0 meV: ln(λ²/m²) = −40.104, the bracket is −46.734, δ₂ = +0.02714 and δₘ = m(0.02714 − 0.05135) = −0.01237 MeV. But m₀ = m − Σ₂(m) = 0.48476 MeV again.
- Checks: δₘ and m − m₀ are masses and δ₂ is a pure number, and all vanish as α → 0. The self-mass is proportional to m and only logarithmic in Λ. The infrared logarithm sits in δ₂, hence in δₘ, never in m − m₀: the counterterm coefficient is not the mass shift.
AnswerAt λ = 1.0 eV, δ₂ = 0.0111 and δₘ = −0.0206 MeV; at λ = 1.0 meV, δ₂ = 0.0271 and δₘ = −0.0124 MeV. In both, m − m₀ = 0.0262 MeV, 5.1 per cent of m, and m₀ = 0.485 MeV.
HardThe muon, 105.658 MeV, and the electron couple to the same photon. With common regulators Λ and λ in Feynman gauge, use the δ₂ quoted in the Medium problem to find Z₂(μ) − Z₂(e) and show it is regulator-free. A student renormalises charge as e = e₀Z₂√Z₃, forgetting Z₁: what charge ratio does that predict for equal bare charges, and what does QED predict? Then add the muon loop to δ₃ at Λ = 1.0 TeV.
- For each species the definition reads q = q₀Z₂√Z₃/Z₁, with that species' own Z₁ and Z₂ but a single Z₃, because every vertex attaches to the same photon field.
- Subtract the quoted δ₂ at the two masses with equal Λ and λ: the ln Λ² and ln λ² terms cancel, leaving Z₂(μ) − Z₂(e) = (3α/4π)ln(mμ²/mₑ²) = (3α/2π)ln 206.768 = 3.4842 × 10⁻³ × 5.3316 = 0.01858. Finite and regulator-free, though each Z₂ alone is neither.
- The student's rule gives qμ/qₑ = Z₂(μ)/Z₂(e) = 1 + 0.0186 to first order, for equal bare charges, a 1.9 per cent mismatch. Muonium 1S–2S spectroscopy finds the two charges equal to a few parts in 10⁹, so the rule fails by about seven orders of magnitude.
- QED's answer: the Ward–Takahashi identity gives Z₁ = Z₂ species by species, so q = √Z₃q₀ for every particle and qμ/qₑ = 1 exactly, to all orders and for any masses. The proton, whose Z₁ and Z₂ carry strong-interaction dressing, is held to the electron the same way: |qₚ + qₑ|/e is measured below 10⁻²¹.
- The muon loop adds −(α/3π)ln(Λ²/mμ²) = −7.743 × 10⁻⁴ × 18.311 = −0.01418 to δ₃, taking it from −0.02243 to −0.03661 and Z₃ from 0.9776 to 0.9634. That rescales e₀ for every species at once and leaves the charge ratio untouched.
- Checks: the Z₂ difference is a pure number that vanishes as mμ → mₑ and as α → 0. It is a difference of unobservable field normalisations, which is how it can reach 1.9 per cent while the charges agree to 10⁻⁹.
AnswerZ₂(μ) − Z₂(e) = 0.0186, free of Λ and λ; forgetting Z₁ would give qμ/qₑ = 1.019, excluded at the 10⁻⁹ level, while QED gives exactly 1. The muon loop moves δ₃ from −0.0224 to −0.0366 for both alike.